Question

A box is kept in the rear of a truck that is travelling
on a level road. The coefficient of sliding friction
between box and truck bed is 1/3. The driver
suddenly applies the brakes, producing a
deceleration g/2. Acceleration of box with respect
to road is:
(1) g/2 forward (2) g/2 backward
(3)g/3 forward (4)g/3 backward

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Answer 1

answer : option (4) g/3 backward.

when driver applies the brake and producing a deceleration g/2, box kept in the rear of the truck try to moves forward [ from law of inertia of motion. ]

due to friction between box and surface of truck, a force applies against relative motion between box and surface.

i.e., F = μmg = 1/3 mg

another force F = ma = m(g/2) acting on box try to moves it forward direction.

so, Fnet = m(g/2) - m(g/3) = m(g/6) acting forward on box with respect to turck.

hence, acceleration of box with respect to road = acceleration of box with respect to truck + acceleration of truck with respect to road

= g/6 + (-g/2)

= -g/3

hence, acceleration of box with respect to road is g/3 backward.

Answer 2

$\huge\bold\purple{Answer:-}$

when driver applies the brake and producing a deceleration g/2, box kept in the rear of the truck try to moves forward [ from law of inertia of motion. ]

due to friction between box and surface of truck, a force applies against relative motion between box and surface.

i.e., F = μmg = 1/3 mg

another force F = ma = m(g/2) acting on box try to moves it forward direction.

so, Fnet = m(g/2) - m(g/3) = m(g/6) acting forward on box with respect to turck.

hence, acceleration of box with respect to road = acceleration of box with respect to truck + acceleration of truck with respect to road

= g/6 + (-g/2)

= -g/3

hence, acceleration of box with respect to road is g/3 backward.